Search
You can find the results of your search below.
Fulltext results:
- Question 1 and 2 Exercise 6.5
- = Suppose events $A$ and $B$ are such that $P(A)=\dfrac{2}{5}, P(B)=\dfrac{2}{5}$ and $P(A \cup B)=\dfrac{1}{2}$.\\ Find $P(A \cap B)$. ====Solution==== We know by addition law of... ), P(B)$ and $P(A \cup B)$, we get $$P(A \cap B)=\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{1}{2}=\dfrac{3}{10}$$
- Question 9 Exercise 6.5
- ncies. The probability of their selection being $\dfrac{1}{7}$ and $\dfrac{1}{5}$ respectively. Find the probability that both will be selected. ====Solution==== \begin{align} P(\text { Ajmal scicction })&=\dfrac{1}{7} \\ \Rightarrow P(\text { Ajmal not selected })&=\dfrac{6}{7} \\ P(\text { Bushra selection })&=\dfrac{1}
- Question 3 & 4 Exercise 6.1
- , Pakistan. =====Question 3(i)===== Prove that $\dfrac{1}{6 !}+\dfrac{2}{7 !}+\dfrac{3}{8 !}=\dfrac{75}{8 !}$ ====Solution==== We are taking the L.H.S of the above given equation. \begin{al
- Question 7 Exercise 6.4
- of getting doublet of an even number is: $$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{3}{36}=\dfrac{1}{12}$$ =====Question 7(ii)===== Two dice are thrown simultaneously. Find the probabilit... ility of getting a number less than 6 is: $$P(B)=\dfrac{n(B)}{n(S)}=\dfrac{10}{36}=\dfrac{5}{18}$$ =====
- Question 9 Exercise 6.3
- omen are: \begin{align}{ }^7 C_4 \cdot{ }^6 C_4&=\dfrac{7 !}{(7-4) ! 4 !} \cdot \dfrac{6 !}{(6-4)}\\\ &= 525\end{align} =====Question 9(ii)===== An $8$-persons... tees are: \begin{align}{ }^7 C_2 \cdot{ }^6 C_6&=\dfrac{7 !}{(7-2) ! 2 !} \cdot \dfrac{6 !}{(6-6) ! 6 !}\\ &=21\end{align} If committee contains $3$ men then i
- Question 1 and 2 Exercise 6.1
- Pakistan. =====Question 1(i)===== Evaluate the $\dfrac{10 !}{3 ! .3 ! \cdot 4 !}$ ====Solution==== \begin{align}\dfrac{10 !}{3 ! \cdot 3 ! \cdot 4 !}&=\dfrac{10.9 .8 \cdot 7 \cdot 6 \cdot 5.4 !}{3 ! \cdot 3 ! \cdot 4 !}\\ &=\dfrac{10.9 .8 .7 .5}{3.2 .1}\\ &=4200 \end{align} ====
- Question 5 Exercise 6.1
- , Pakistan. =====Question 5(i)===== Show that: $\dfrac{(2 n) !}{n !}=2^n(1.3 .5 \ldots(2 n-1))$ ====Solu... taking L.H.S of the above equation \begin{align}\dfrac{(2 n) !}{n !}&=\dfrac{1}{n !}[(2 n)(2 n-1)(2 n-2) \\ &=(2 n-3)(2 n-4)(2 n-5) \ldots(2 n-(2 n-4))\\ &(2 n... bove equation are total $2 n$ terms \begin{align}\dfrac{(2 n) !}{n !}&=\dfrac{1}{n !}[(2 n)(2 n-1)(2 n-2)
- Question 11 Review Exercise 6
- $$ P(orange) The orange color covers one fourth $\dfrac{1}{4}$ of the spinner, thus the probability of is: $$\quad P( orange )=\dfrac{1}{4}$$ P(Red or Green) Red color cover one fourth $\dfrac{1}{4}$ and green color ccvers one fourth $\dfrac{1}{4}$ of the spinner. Therefore, \begin{align}P(\oper
- Question 1 and 2 Exercise 6.2
- e $^6 P_6$ ====Solution==== \begin{align}^6 P_6&=\dfrac{6 !}{(6-6) !}\\ &=6 !=720\end{align} =====Questi... 0} P_2$ ====Solution==== \begin{align}^{20} P_2&=\dfrac{20 !}{(20-2) !}\\ &=\dfrac{20.19 .18 !}{18 !}\\ &=20 \times 19=380\end{align} =====Question 1(iii)===== ... 6} P_3$ ====Solution==== \begin{align}^{16} P_3&=\dfrac{16 !}{(16-3) ! }\\ &=\dfrac{16 \cdot 15 \cdot 14
- Question 2 Review Exercise 6
- { }^{2 n} C_r&={ }^{2 n} C_{r+2} \\ \Rightarrow \dfrac{(2 n) !}{(2 n-r) ! r !}&=\dfrac{(2 n) !}{(2 n-(r+2)) !(r+2) !}\end{align} Dividing both sides by $(2 n)$ ! we get \begin{align} \Rightarrow \dfrac{1}{(2 n-r) ! r !}&=\dfrac{1}{(2 n-r-2) !(r+2) !} \\ \Rightarrow \dfrac{1}{(2 n-r)(2 n-r-1)(2 n-r-2) ! r
- Question 3 and 4 Exercise 6.2
- f the equation \begin{align}n(^{n-1} P_{r-1})&=n \dfrac{(n-1) !}{((n-1)-(r-1)) !} \\ & =\dfrac{n(n-1) !}{(n-r) !}\\ &=\dfrac{n !}{(n-r) !}\\ &=^n P_r\end{align} =====Question 3(ii)===== Prove by Fun... ation \begin{align}^{n-1} P_r+r(^{n-1} P_{r-1})&=\dfrac{(n-1) !}{(n-1-r) !}+r \dfrac{(n-1) !}{(n-1-(r-1))
- Question 4 Exercise 6.3
- S \begin{align} { }^{n-1} C_r+{ }^{n-1} C_{r-1}&=\dfrac{(n-1) !}{(n-r-1) ! r !}+\dfrac{(n-1) !}{(n-1-(r-1)) !(r-1) !} \\ & =\dfrac{(n-1) !}{(n-r-1) ! r(r-1) !}+\dfrac{(n-1) !}{(n-r) !(r-1) !} \\ & =\dfrac{(n-1) !}{(n-r-1) ! r(r-1) !}+\
- Question 3 Exercise 6.4
- bility for each individual outcome to occur is $$\dfrac{1}{256}$$ $8$ answers are correct. Let $$A=\{8\}... stion to be correct by one way i.e. $${ }^8 C_8=\dfrac{8 !}{(8-8) ! 8 !}=1$$ Therefore probability to $8$ answers are correct is: $$P(A)=\dfrac{1}{256}$$ =====Question 3(b)===== A true or fal... bility for each individual outcome to occur is $$\dfrac{1}{256}$$ $7$ answers are correct Let $$B=\{7\}$
- Question 10 Exercise 6.5
- \begin{align}n(S)&={ }^{30} C_2\\ &=435\\ P(A)&=\dfrac{^{20} C_2}{^{30} C_2}\\ &=\dfrac{190}{435}=\dfrac{38}{87}\\ P(B)&=\dfrac{^{22} C_2}{^{30} C_2}\\ &=\dfrac{231}{435}=\dfrac{77}{145}\end{align} Also probab
- Question 1 Exercise 6.3
- e given: \begin{align}&^n C_2=36\\ & \Rightarrow \dfrac{n !}{(n-2) ! 2 !}=36 \\ & \Rightarrow \dfrac{n(n-1)(n-2) !}{(n-2) ! \cdot 2}=36 \\ & \Rightarrow n(n-1)=... } & { }^{n+1} C_4=6 .^{n-1} C_2 \\ & \Rightarrow \dfrac{(n+1) !}{(n+1-4) ! 4 !}=6\dfrac{(n-1) !}{(n-1-2) ! 2 !} \\ & \Rightarrow \dfrac{(n+1) n(n-1) !}{(n-3) !